What is the integral from 0 to 1 of 1/(e^{3x)} ?

The integral from 0 to 1 of 1/(e^{3x)} is (e^3-1)/(3e^3)

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integral from 0 to 1 of 1/(e^{3x)}

\int_{0}^{1} \frac{1}{e ^{3 x}} d x

\frac{e ^{3} - 1}{3 e ^{3}}

Decimal Notation

0.31673 \dots

Solution steps

\int_{0}^{1} \frac{1}{e ^{3 x}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{3 x}} = e^{- 3 x}

= \int_{0}^{1} e^{- 3 x} d x

Apply u-substitution

: \int_{0}^{- 3} - \frac{1}{3} e^{u} d u

= \int_{0}^{- 3} - \frac{1}{3} e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 3}^{0} - \frac{1}{3} e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \frac{1}{3} \cdot \int_{- 3}^{0} e^{u} d u)

Use the common integral:

\int e^{u} d u = e^{u}

= - (- \frac{1}{3} [e^{u}]_{- 3}^{0})

Simplify

= \frac{1}{3} [e^{u}]_{- 3}^{0}

Compute the boundaries:

1 - \frac{1}{e ^{3}}

= \frac{1}{3} (1 - \frac{1}{e ^{3}})

Simplify

= \frac{e ^{3} - 1}{3 e ^{3}}

\frac{d}{d x} (e^{- a x^{2}})

n = 0 \sum \infty \frac{e ^{n}}{π ^{n}}

\frac{d}{d x} (3 (4 e^{2 x} x + 4 e^{2 x}))

\int \frac{x}{10 - x ^{2}} d x

n = 1 \sum \infty \frac{1}{( 2 n - 1 )}

What is the integral from 0 to 1 of 1/(e^{3x)} ?
The integral from 0 to 1 of 1/(e^{3x)} is (e^3-1)/(3e^3)